Python：按键的前半部分/部分重新排序字典

Question

I need to reorder a very large OrderedDict by the first 'CSV' of its keys. For example I have a dictionary like this:

 a = {'40,70': AAAAAA, '0,12': XXXXXXXX, '20,38': YYYYY}

I need it to be ordered on the first number of the key which is always two numbers separated by a comma (I need the second number to then be the ordering factor). The values or Letters are currently of no importance. I need it to be like this:

b = {'0,12': XXXXXXXX, '20,38': YYYYY, '40,70': AAAAAA}

The dictionary is too big to split each key in a loop - Is there a faster way of doing this? I have tried to make this as clear as possible.

Thanks.

Answer 1

you can use ordered dict to reorder the dict

from collections import OrderedDict  
OrderedDict(sorted(a.items(), key=lambda ele:[int(item) for item in ele[0].split(',')])) 

Answer 2

best way would be to create tuple keys in your dictionary, then sort using natural order.

If you want to keep the keys as strings, then sort the items of the dictionary into a tuple, with a key function converting the key as a tuple of integers. This provides a tiebreaker in case 2 first numbers are equal:

a = {'40,70': "AAAAAA", '0,12': "XXXXXXXX", '0,10': "XXXXXXXX", '20,38': "YYYYY"}

import collections

b = collections.OrderedDict(sorted(a.items(),key = lambda e : tuple(map(int,e[0].split(",")))))

print(b)

result:

OrderedDict([('0,10', 'XXXXXXXX'), ('0,12', 'XXXXXXXX'), ('20,38', 'YYYYY'), ('40,70', 'AAAAAA')])

Answer 3

第二个号码作为决胜局：

ordered_dict = dict(sorted(a.items(), key=lambda x: [int(x[0].split(",")[0]), int(x[0].split(",")[1])]))

Answer 4

I think using OrderedDict you can use something like this. I am not sure if this is the fastest way.

To sort numerically, I assumed you will always get an integer before comma.

To use second part as second key, and turned the key to a tuple.

from collections import OrderedDict
a = {'40,70': 'AAAAAA', '0,12': 'XXXXXXXX', '20,38': 'YYYYY', '0,38': 'AAYY'}
b = OrderedDict(sorted(a.items(), key=lambda x: (int(x[0].split(',')[0]), int(x[0].split(',')[1]))))
print(b)

result is

OrderedDict([('0,12', 'XXXXXXXX'), ('0,38', 'AAYY'), ('20,38', 'YYYYY'), ('40,70', 'AAAAAA')])

Answer 5

Use OrderedDict

from collections import OrderedDict

a = {'40,70': 'AAAAAA', '0,12': 'XXXXXXXX', '20,38': 'YYYYY'}

b = OrderedDict(sorted(a.items()))